Question

From the foot of an inclined plane, whose angle is tan (1/3), a shot is projected with a velocity of 10 m/s at an angle of 60° with the horizontal up the plane. Find the time of flight and range. (g = 9.8 m/s)

Answer 1

Explanation:

in this way we can solve the above problem my friend please check your answers

Answer 2

Answer:20/3

Explanation:

theta= the angle of body from horizontal = 60°

Alpha= Angle of inclination 30°

The formula for range of a projectile on inclined plane is

V^2[sin(2theta-alpha) - sin(alpha)/gcos^2(alpha)]

[((10)^2×sin(120-30) - sin(30))/(9.8×3/2)]

Putting in the values we get-

(100×1 - 1/2)/4.9×3

Which will roughly be equal to 20/3

Hope it helps  :)